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>> No.11638449 [View]
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11638449

>>11638166
You can actually go a step further. Assuming a circular shaft, you can show that angular displacement as a function of position and time is governed by [math] \rho\theta_{tt}=G\theta_{xx} [/math], where the subscripts are partial derivatives. This is just the wave equation, and it can also be shown that [math] c=\sqrt{G/\rho} [/math] is the speed at which this torsional wave propagates down the shaft.
Roughly, steel has shear modulus of ~80 GPa and density of ~8000 kg/m^3. So the wave propagates at sqrt(10000000) meters per second. Time to cover a lightyear or 9.4E15 meters is then ~3*10^12 seconds. This is assuming a circular shaft, of course. Pretty interesting that the radius doesn't matter, IMO.

>> No.11561086 [View]
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11561086

>>11560509
>>11560513
Not necessarily, but the theory of closures is a go.
A is the non-measurable set. We assume it's bounded for convenience. For some set [math]A[/math] and any [math]n \in \mathbb{N}[/math], there is an [math]S_n[/math] measurable with [math]A \subseteq S_n[/math] such that [math]\mu(S_n) < inf_{B ~ measurable ~ such ~ that ~ A \subset B} \mu (B) + 2^{-n}[/math], and the intersection [math]S = \cap_n S_n[/math] is a measurable set(since it's a countable intersection of measurable sets), contains [math]A[/math] and has measure [math]\mu(S) = inf_{B ~ measurable ~ such ~ that ~ A \subset B} \mu (B)[/math], and thus should be the "closure" you want.
Finally, for any other [math]A \subseteq C[/math] with [math]C[/math] measurable, it follows from the definition that [math]\mu(C \cap S) \geq inf_{B ~ measurable ~ such ~ that ~ A \subset B} \mu(B) = \mu(S)[/math], which implies that either [math]\mu (C -S) > 0[/math] or that they are the same except for a set o measure zero.
You then globalize the function through basic axiom of choice abuse.
>>11560635
Impulse is just change in momentum. Does B face a greater change in momentum than A?
>>11561039
Only if you have net charge.

>> No.11561012 [DELETED]  [View]
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11561012

>>11560509
>>11560513
For some set [math] A [/math] and any [math] n \in \mathbb{N} [/math], there is an [math] S_n \in \mathcal{L} [/math] such that [math] \mu(S_n) < inf_{B ~ measurable ~ such ~ that ~ A \subset B} \mu (B) + 2^{-n} [/math], and the intersection [math] S = \cap_n S_n [/math] has measure [math] \mu(S) = inf_{B ~ measurable ~ such ~ that ~ A \subset B} \mu (B) [/math], and thus should be what you want. Finally, for any other [math] A \subseteq C [/math] with [math] C [/math] measurable, it follows from the definition that [math] \mu(C \cap S) \geq inf_{B ~ measurable ~ such ~ that ~ A \subset B} \mu(B) = \mu(S) [/math], which implies that either [math] \mu (C -S) > 0 [/math] or that they are the same except for a set to measure zero.

>>11560635
Impulse is just change in momentum. Does B have a greater change in momentum than A?

>> No.11534558 [View]
File: 120 KB, 648x800, 783dac73602b4db969d9f618ff1e1975.jpg [View same] [iqdb] [saucenao] [google]
11534558

>>11534044
x proportional to y and inversely prop. to z means [math] x=k\frac{y}{z} [/math] with k being a constant of proportionality. [math] 6=k\frac{10}{15}\implies k=9 [/math]. So [math] 92=9\times\frac{107}{z}\implies z=963/92 [/math].
>>11528997
[math] v\approx\sqrt{2P/\rho} [/math]. Velocity, gauge pressure and density, respectively.
>>11525657
E definitely does increase as the plates are brought together (and voltage is maintained). Remember that Gauss's law only applies to closed surfaces.

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